Q 11-03-053JEE MainJEE Main 2025 (23 Jan, Shift 1)Medium
Two particles are located at equal distance from origin. The position vectors of those are represented by $\vec A = 2\hat i + 3n\hat j + 2\hat k$ and $\vec B = 2\hat i - 2\hat j + 4p\hat k$, respectively. If both the vectors are at right angle to each other, the value of $n^{-1}$ is ______.
Numerical value type. Enter your answer.
Answer: 3
Equal distances from the origin: $|\vec A| = |\vec B|$
$$4 + 9n^2 + 4 = 4 + 4 + 16p^2 \Rightarrow 9n^2 = 16p^2 \Rightarrow 3n = \pm4p$$
Perpendicular: $\vec A\cdot\vec B = 0$
$$4 - 6n + 8p = 0 \Rightarrow 3n = 2 + 4p$$
$3n = 4p$ would give $0 = 2$, impossible. So $3n = -4p$:
$$-4p = 2 + 4p \Rightarrow p = -\frac{1}{4}, \quad n = \frac{1}{3}$$
$n^{-1} = 3$
Solution by Sreeraj P, M.Sc Physics